Answer:
a=1 converges for 0<x<8
Step-by-step explanation:
Given that a geometric series infinite is given in summation form as
[tex]\Sigma _0^\infty (\frac{x-4}{4} )^n[/tex]
Here I term is when n =0
So a= I term = [tex](\frac{x-4}{4} )^0=1[/tex]
r = common ratio = [tex]\frac{x-4}{4}[/tex]
This series converges only if
[tex]|r|<1[/tex]
[tex]|\frac{x-4}{4}|<1\\-4<x-4<4\\0<x<8[/tex]
For 0<x<8, we get absolute value of r is less than 1.
Hence series converges for [tex]0<x<8[/tex]
